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There is an epistatic gene that is in control of pigment that is deposited on the...

There is an epistatic gene that is in control of pigment that is deposited on the hair of a few mice. One dominant epistatic gene E for the color to show or the mouse is white. This is a dihybrid cross.

1. Create a punnet square for a dihybrid cross of two mice both which are heterozygotic for their color and epistatic gene.

2. Which percent of these will be white?

3. This is different than Mendel's F2 gen with dihybrid cross pea plants, why?

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Answer #1

Hair colour of mouse :-
Trait 1 -
Alleles - B (Dominant) and b (recessive).
BB and Bb = Black hairs.
bb = Brown hairs.

Trait 2 -
Alleles - E ( Dominant) and e (recessive).
EE and Ee = Normal hair color (According to B and b).
ee = White hairs ( Epistasis).

Cross - BbEe × BbEe.

Gametes types - BE, Be, bE and be.

Punnett square -

Result :-
9. Black = 1. BBEE + 2. BbEE + 2. BBEe + 4. BbEe

3. Brown = 1. bbEE + 2. bbEe

4. White = 1. BBee + 2. Bbee + 1. bbee

Phenotype ratio = Black : Brown : White = 9:3:4.

Difference from mendel's dihybrid cross is of phenotype ratio.
Result of mendel's dihybrid cross = 9:3:3:1.
Result of this case = 9:3:4.

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